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2 years ago

7

Complete each congruency statement and name the rule used. If you cannot show the triangles are congruent from the given informa

tion, leave the triangle's name blank and write CNBD for "Cannot be determined" in place of the rule.

1 answer:

Tpy6a [65]2 years ago

5 0

Answer:

<u>The first picture</u>

ΔBOY ≅ ΔMAN by ASA

Two angles and the included side of ΔBOY are congruent to corresponding parts of ΔMAN

<u>The second picture</u>

ΔBLK ≅ ____ by CNBD

We can see parallel segments and transversals forming 3 congruent angles in both triangles but this is not enough to state the triangles are congruent.

BA ∩ LC = K is not a midsegment.

With the info we have we can only state that the triangles are similar but not congruent.

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a manufacturer has determined that if his company sells x items per day then their profit is given by P=-x^2+ 700 x-10000 euros.

Radda [10]

Answer:

15

Step-by-step explanation:

To find the minimum amount of items, x, that need to be sold for the manufacturer to make a profit, we can use the quadratic formula

$x = \frac{-b+\sqrt{b^{2}-4ac } }{2a}; x= \frac{-b-\sqrt{b^{2}-4ac } }{2a}$

$x= \frac{-700+\sqrt{700^{2}-4(a)(-10000) } }{2(1)}; x= \frac{-700-\sqrt{700^{2}-4(a)(-10000) } }{2(1)}$

$x= \frac{-700+\sqrt{530000} }{2}; x= \frac{-700-\sqrt{530000} }{2}$

$x=\frac{-700+100\sqrt{53} }{2}; x=\frac{-700-100\sqrt{53} }{2};$

$x=-350 + 50\sqrt53 = 14.005; x = -350 - 50\sqrt53 = -714.005$

In context of the problem, we can only rely on the positive value, as the negative value would lead to a loss of profit.

$(-714.005)^2+700(-714.005)-10000=-0.360$

Furthermore, we must round to the nearest whole number, as you cannot make part of an item.

Lastly, if you were to plug in 14 into the equation for Profit, you would still have a negative number (i.e. a negative profit), thus requiring the manufacturer to make no less than 15 items to make a profit:

$(14)^2+700(14)-10000=-4\\\\(15)^2+700(15)-10000=725$

3 0

1 year ago

What is the arc length S what is shown below?

Solnce55 [7]

Answer:

14 cm

Step-by-step explanation:

The applicable formula is ...

s = rθ

s = (7 cm)(2) = 14 cm

3 0

2 years ago

Use the zero product property to find the solution to the equation x^2 +12=7x

pochemuha

X2 + 12 = 7x x2 – 7x + 12 = 0factor the equation (x – 3 ) ( x – 4 ) = 0 Then equating both factor to 0 X – 3 = 0 X = 3 And X – 4 = 0 X = 4 So the answer is X = 3 X =4

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3 years ago

Read 2 more answers

Need help with this question

photoshop1234 [79]

$y = 3/4 x + ( - 2)$

4 0

2 years ago

Read 2 more answers

Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T

frosja888 [35]

Answer:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

$p_v =P(Z$

a) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

b) $\chi^2 =\frac{10-1}{1.96} 4 =18.367$

$p_v =P(\chi^2 >18.367)=0.0311$

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation

$\bar X=7.25$ represent the sample mean

$s=1.2$ represent the sample standard deviation

$\sigma=1.4$ represent the population standard deviation

$n=10$ sample size

$\mu_o =7.5$ represent the value that we want to test

$\alpha=0.05,0.01$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:

Null hypothesis: $\mu \geq 7.5$

Alternative hypothesis: $\mu < 7.5$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565$

4)P-value

Since is a left tailed test the p value would be:

$p_v =P(Z$

5) Conclusion

Part a

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

$n=10$ represent the sample size

$\alpha=0.01$ represent the confidence level

$s^2 =4$ represent the sample variance obtained

$\sigma^2_0 =1.96$ represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \leq 1.96$

Alternative hypothesis: $\sigma^2 >1.96$

Calculate the statistic

For this test we can use the following statistic:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

$\chi^2 =\frac{10-1}{1.96} 4 =18.367$

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

$p_v =P(\chi^2 >18.367)=0.0311$

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that $p_v >\alpha$ so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

4 0

3 years ago